Mean Value Theorem Problem?

Question

Which of the following functions
A. f(x)=1/(x-1) on [0,2]
B. f(x)=x^1/3 on [0,1]
C. f(x)=|x| on [0,1]
satisfy the hypotheses of the MVT?

I put only B but I got that wrong. It can be two or more answers... thank you!!

Answer 1

The answer is B and C. A is out because the original function is undefined at x = 1 which is on [0,2]. B is in because it is continuous for all values of x, it is not differentiable (the derivative being continuous) at x = 0, but we are only checking differentiablility on (0,2), which means only checking in between 0 and 2, but not the actual values of 0 and 2. C is also in because the functions is continuous for all values of x, the derivative is discontinuous at 0 (because of the sharp point on f(x) at 0), but we are only checking differentiability on (0,2). Hope that helps.

Answer 2

I forget what the hypotheses exactly are, but I imagine it would suffice to be differentiable on the interior of the interval and continuous over the whole interval. And C. satisfies both those criteria, since on (0,1] it's just the same thing in every sense as g(x)=x (and it's continuous at 0).

Answer 3

C does, I think. (f(b)-f(a))/(b-a) will be constant (1) for abs(x) (which is the same as y=x over [0,1]).